Solving the equation found at 3:40 using a half interval search: Let the linear dimensions be in m and areas in m². Then the equation becomes √(a² - 36) = √(a² - 1089) + √(a² - 1521). Let Δ = √(a² - 36) - (√(a² - 1089) + √(a² - 1521)), the difference between the two sides of the equation. We wish to find a value of a that will make that difference Δ as small as possible. We note that a cannot be less than 39 or a² - 1521 would be negative, producing an imaginary number for √(a² - 1521). For a = 39, Δ is positive. Experimentally, I find Δ is negative for a = 55. So, I write a computer program to perform a half interval search for a value of a which minimizes Δ, using search range limits of a = 39 (low end) and a = 55 (high end). I compute Δ for the midpoint, 47. If a negative result, I check 43 (half way between 39 and 47) otherwise 51 (half way between 47 and 55). I continue to zoom in by cutting the distance in half each time. My computer result is 42.000000000000 to within the precision of its calculations. Now, is 42 exact? I substitute a = 42 into the original equation and get √(42² - 36) = √(42² = 1089) + √(42² - 1521), which simplifies to √(1728) = √(675) + √(243). I square both sides: 1728 = 675 + 2(√(675))(√(243)) + 243, which simplifies to 1728 = 918 + 2(√(164025)), 1728 = 918 + 2(405), and 1728 = 1728. So, a = 42 is an exact answer. Here is my computer program, written in python, and its result: import math a = 47.0 da = 4.0 # define function to compute delta = ∆ def compute_delta(a): delta = math.sqrt(a*a - 36) - (math.sqrt(a*a - 1089) + math.sqrt(a*a - 1521)) return(delta) for n in range(0,51): # after 50 divisions by 2, da is sufficiently small delta = compute_delta(a) if (delta > 0): a = a + da else: a = a - da da = da/2 a = round(a , 12) # round a to 9 decimal places formatted = "{:.12f}".format(a) print(f"a = ",formatted) Output: a = 42.000000000000 === Code Execution Successful ===
Excellent! You can carry this further to show that a = 42 exactly. You found tan(α) = (33√3)/45. You do not need to compute α. Construct a right triangle with side 33√3 opposite angle α and 45 adjacent. The hypotenuse squared is (33√3)² + (45)² = 3267 + 2025 = 5292. Hypotenuse = √(5292) = 42√3. sin(α) = (33√3)/(42√3) = 33/42. On the second line, a = 33/(sin(α)) = 33/(33/42) = 42 exactly.
Let CDE be the equilateral triangle and move AB horizontally to the right until E be on line segment AB. Let x=120-y BE=EC*cos x ->33=a*cos(120-y) …….(1) AE=ED*cos y->6=a*cos y ……..(2) Dividing (1) by (2) 11/2=cos(120-y)/cos y =(cos 120*cos y+sin 120*sin y)/cos y=-1/2+sqrt(3)/2*tan y tan y=4*sqrt(3) from which sec y=sqrt(tan(y)^2+1)=7 ->cos y=1/7 replacing to (2) 6=a*1/7 ->a=42 Area CÈDE=sqrt(3)/4*a^2=441*sqrt(3)
Solving the equation found at 3:40 using a half interval search: Let the linear dimensions be in m and areas in m². Then the equation becomes √(a² - 36) = √(a² - 1089) + √(a² - 1521). Let Δ = √(a² - 36) - (√(a² - 1089) + √(a² - 1521)), the difference between the two sides of the equation. We wish to find a value of a that will make that difference Δ as small as possible. We note that a cannot be less than 39 or a² - 1521 would be negative, producing an imaginary number for √(a² - 1521). For a = 39, Δ is positive. Experimentally, I find Δ is negative for a = 55. So, I write a computer program to perform a half interval search for a value of a which minimizes Δ, using search range limits of a = 39 (low end) and a = 55 (high end). I compute Δ for the midpoint, 47. If a negative result, I check 43 (half way between 39 and 47) otherwise 51 (half way between 47 and 55). I continue to zoom in by cutting the distance in half each time. My computer result is 42.000000000000 to within the precision of its calculations. Now, is 42 exact? I substitute a = 42 into the original equation and get √(42² - 36) = √(42² = 1089) + √(42² - 1521), which simplifies to √(1728) = √(675) + √(243). I square both sides: 1728 = 675 + 2(√(675))(√(243)) + 243, which simplifies to 1728 = 918 + 2(√(164025)), 1728 = 918 + 2(405), and 1728 = 1728. So, a = 42 is an exact answer.
Here is my computer program, written in python, and its result:
import math
a = 47.0
da = 4.0
# define function to compute delta = ∆
def compute_delta(a):
delta = math.sqrt(a*a - 36) - (math.sqrt(a*a - 1089) + math.sqrt(a*a - 1521))
return(delta)
for n in range(0,51): # after 50 divisions by 2, da is sufficiently small
delta = compute_delta(a)
if (delta > 0):
a = a + da
else:
a = a - da
da = da/2
a = round(a , 12) # round a to 9 decimal places
formatted = "{:.12f}".format(a)
print(f"a = ",formatted)
Output:
a = 42.000000000000
=== Code Execution Successful ===
We call the triangle ECD and let H be the perpendicular projection of point E on BC. We put
α+β= 60°
a = 33/sinα ; a = 6/sinβ
Equalling:
33 sinβ = 6 sinα
sin(60°-α) = 6/33 sinα
½(√3cosα - sinα) = 6/33 sinα
√3/tanα - 1 = 12/33
√3/tanα = 45/33
tanα= 33√3/45 --> α=51,787°
Triangle side:
a = 33/sinα = 42 m
Area of equilateral triangle:
A = √3/4 a² = 441√3 m²
A = 763,83 m² ( Solved √ )
Excellent! You can carry this further to show that a = 42 exactly. You found tan(α) = (33√3)/45. You do not need to compute α. Construct a right triangle with side 33√3 opposite angle α and 45 adjacent. The hypotenuse squared is (33√3)² + (45)² = 3267 + 2025 = 5292. Hypotenuse = √(5292) = 42√3. sin(α) = (33√3)/(42√3) = 33/42. On the second line, a = 33/(sin(α)) = 33/(33/42) = 42 exactly.
@@jimlocke9320
Excelent !!!
Let CDE be the equilateral triangle and move AB horizontally to the right until E be on line segment AB.
Let x=120-y
BE=EC*cos x ->33=a*cos(120-y) …….(1)
AE=ED*cos y->6=a*cos y ……..(2)
Dividing (1) by (2)
11/2=cos(120-y)/cos y =(cos 120*cos y+sin 120*sin y)/cos y=-1/2+sqrt(3)/2*tan y
tan y=4*sqrt(3) from which sec y=sqrt(tan(y)^2+1)=7 ->cos y=1/7 replacing to (2)
6=a*1/7 ->a=42
Area CÈDE=sqrt(3)/4*a^2=441*sqrt(3)
(6m)^2 (33m)^2={36m^2+1089m^2}=1125m^4 {60°A+60°B+60°C}=180°ABC 1125m^4/180°ABC=10013.1 10^1013.1 2^5^2^53. 1^1^2^13. 23. (ABC ➖ 3ABC+2).